Pandas：获取系列的前 10 个元素
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Pandas: get first 10 elements of a series
提问by chintan s
I have a data frame with a column tfidf_sortedas follows:
我有一个带有列的数据框，tfidf_sorted如下所示：
tfidf_sorted
0 [(morrell, 45.9736796), (football, 25.58352014...
1 [(melatonin, 48.0010051405), (lewy, 27.5842077...
2 [(blues, 36.5746634797), (harpdog, 20.58669641...
3 [(lem, 35.1570832476), (rottensteiner, 30.8800...
4 [(genka, 51.4667410433), (legendaarne, 30.8800...
The type(df.tfidf_sorted)returns pandas.core.series.Series.
的type(df.tfidf_sorted)回报pandas.core.series.Series。
This column was created as follows:
此列的创建方式如下：
df['tfidf_sorted'] = df['tfidf'].apply(lambda y: sorted(y.items(), key=lambda x: x[1], reverse=True))
where tfidfis a dictionary.
tfidf字典在哪里。
How do I get the first 10 key-value pairs from tfidf_sorted?
如何从 中获取前 10 个键值对tfidf_sorted？
采纳答案by jezrael
IIUC you can use:
您可以使用 IIUC：
from itertools import chain
#flat nested lists
a = list(chain.from_iterable(df['tfidf_sorted']))
#sorting
a.sort(key=lambda x: x[1], reverse=True)
#get 10 top
print (a[:10])
Or if need top 10 per row add [:10]:
或者，如果需要每行前 10 个添加[:10]：
df['tfidf_sorted'] = df['tfidf'].apply(lambda y: (sorted(y.items(), key=lambda x: x[1], reverse=True))[:10])
